This concerns a number of basic questions about ample line bundles on a variety $X$
and maps to projective space. I have searched related questions and not found answers, but I apologize if I missed something. I'll work with schemes of finite type over a field $k$ for simplicity.

Background

A quasi-coherent sheaf $F$ on a $k$-scheme $X$ is globally generated if the natural map $H^{0}(X,F)\otimes \mathcal{O}_{X} \rightarrow F$ is a surjection of sheaves. Basically, this says that for any point on $X$, there is at least one section of $F$ that doesn't vanish at that point, so there are enough sections of $F$ to see all the points of $X$. (EDIT: As pointed out in the comments below, this last sentence does not describe a situation equivalent to being globally generated. Perhaps it is better to say that
globally generated means that for each point $x \in X$, $F$ has some rank $r$ at $x$ and globally generated means that there are at least $r$ sections of $F$ that are linearly independent over $x$.)

The notion of globally generated is especially useful when $F=L$ is a line bundle on $X$.
If $V$ is a finite dimensional subspace of $H^{0}(X,L)$ such that $V \otimes \mathcal{O} \rightarrow L$ is surjective, then we get a morphism $\varphi_{V}:X \rightarrow \mathbb{P}(V)$
by the universal property of the projective space $\mathbb{P}(V)$ of hyperplanes in $V$.
Essentially, given a point $x \in X$, we look at the fibre over $x$ of the surjection
$V \otimes \mathcal{O} \rightarrow L$ to get a quotient $V \rightarrow L_{x}$. The kernel
is a hyperplane in $V$, and the morphism $\varphi_{V}$ sends $x$ to that hyperplane as a point in $\mathbb{P}(V)$.

So how to build globally generated sheaves? A line bundle $L$ is called ample
if for every coherent sheaf $F$, $F \otimes L^{\otimes n}$ is globally generated for all large $n$. The smallest $n$ after which this becomes true can depend on $F$.

Finally, a line bundle is called very ample if $L$ is globally generated and $\varphi_{V}$ is an embedding
for some subspace of sections $V$.

There are various properties of and criteria for ample line bundles, which can be found in Hartshorne, for example. What we need for the below questions are the following: $L$ is ample if and only if $L^{m}$ is ample for some $m$ if and only if $L^{n}$
is very ample for some $n$; if $L$ is ample, eventually $L^{k}$ will have sections, be globally generated, be very ample, and have no higher cohomology.

Questions

Are there simple examples (say on a curve or surface) of line bundles that are globally generated but not ample, of ample line bundles with no sections, of ample line bundles that are globally generated but not very ample, and of very ample line bundles with higher cohomology?

Given an ample line bundle $L$, what is the minimal number $k$ so that I can be sure $L^{k}$ has sections, is globally generated, is very ample? Is $k$ related to the dimension of $X$?

If $L$ is very ample, I can use it to embed $X$ into some projective space. Then by projecting from points off of $X \subset \mathbb{P}^{N}$, I can eventually get a finite morphism $X \rightarrow \mathbb{P}^{d}$, where $d$ is the dimension of $X$. But what if
I just know that $L$ is ample and globally generated? Can I also use it to get such a finite morphism to $\mathbb{P}^{d}$?

Your claim in the second paragraph (Basically...) is wrong. This is only true for line bundles, since any nonzero number generates $\mathbb{C}$ as a vector space.
–
Andrea FerrettiMay 22 '10 at 22:05

2

The structure sheaf is globally generated, but is not ample if $X$ is projective of positive dimension.
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EmertonMay 22 '10 at 23:20

I don't follow Andrea's comment above. Maybe I am missing something, but if $H^{0}(X,F)\otimes \mathcal{O} \rightarrow F \rightarrow 0$ is exact, then it is also exact on fibres and hence the restriction of sections to a point is surjective on fibres. If the fibre of $F$ at a point is zero, then of course any section will vanish there. But so long as $F$ is supported everywhere (even if not a line bundle) and globally generated, won't there be a section not vanishing at a chosen point?
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A. PascalMay 23 '10 at 23:32

1

Yes, but this is weaker than the statement that $F$ is globally generated; the latter means that at each point there are enough sections to span the whole stalk (which will take more than a single non-vanishing section at that point, if $F$ has rank $> 1$ in a n.h. of that point.
–
EmertonMay 24 '10 at 2:01

4 Answers
4

1. Are there simple examples (say on a curve or surface) of line bundles that are globally generated but not ample, of ample line bundles with no sections, of ample line bundles that are globally generated but not very ample, and of very ample line bundles with higher cohomology?

On a curve of genus $g$, a general divisor of degree $d \le g-1$ has no sections. Of course, if $d>0$ then it is ample.

$K_X$ on a hyperelliptic curve is globally generated but not very ample.

you see that $H^1(\mathcal O_C(1))=H^2(\mathcal O_{\mathbb P^2}(1-d))$ which is dual to $H^0(\mathcal O_{\mathbb P^2}(d-4))$. So that's nonzero for $\ge4$.

2. Given an ample line bundle $L$, what is the minimal number $k$ so that I can be sure $L^k$ has sections, is globally generated, is very ample? Is $k$ related to the dimension of $X$?

Again, just look at the divisor of a degree 1 on a curve of genus $g$. You need $k\ge g$, so you see that there is no bound in terms of the dimension.

It turns out that a better right question to ask is about the adjoint line bundles $\omega_X\otimes L^k$ ($K_X+kL$ written additively). Then the basic guiding conjecture is by Fujita, and which says that for $k\ge \dim X+1$ the sheaf is globally generated, and for $k\ge \dim X+2$ it is very ample. This is proved for $\dim X=2$, proved with slightly worse bounds for $\dim X=3$. For higher dimensions the best result is due to Angehrn-Siu who gave a quadratic bound on $k$ instead of linear. There are some small improvements for some special cases.

3. If $L$ is very ample, I can use it to embed $X$ into some projective space. Then by projecting from points off of $X\subset \mathbb P^N$, I can eventually get a finite morphism $X\to \mathbb{P}^d$, where $d$ is the dimension of $X$. But what if I just know that $L$ is ample and globally generated? Can I also use it to get such a finite morphism to $\mathbb P^d$?

But of course $L$ gives a morphism $f$, and it follows that $f$ is finite: $f$ contacts no curve so $f$ is quasifinite, and $f$ is projective (since $X$ was assumed to be projective). And quasifinite + proper = finite.

A comment and question on the the answer to 3. I suppose that the reason the morphism cannot contract a curve is that then the divisor class associated to $L$ (pulled back from the hyperplane on projective space) and all powers of it would have to intersect the contracted curve trivially. But then by Nakai-Moishezon, the divisor would have to be trivial. Is this correct? Now why do we have quasifinite + proper = finite?
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A. PascalMay 23 '10 at 13:11

@A.Pascal: if $H=L^n$ is a hyperplane section then $H.C=deg(C)>0$, so for an ample line bundle you always have $L.C>0$ just by this. Quasifinite+proper = finite: see Hartshorne.
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VA.May 24 '10 at 1:09

The answers already given seem to settle your questions, but I'd like to emphasise one point that hasn't been mentioned.

Suppose X is a projective variety, and $\pi: X' \rightarrow X$ any morphism which contracts at least one curve in X'. (For example, it could be the blowup of any subvariety of X of codimension at least 2.) Then for any very ample line bundle L on X, the pullback $\pi^*L$ is globally generated but not ample, because it has degree zero on any curve contracted by $\pi$. (So to get a simple example for the first part of your first question, let $\pi$ be the blowup of P^2 in a point, and L the class of a line.)

The point I want to make is that this is not just a class of examples, but in fact the general case. To see this, suppose X is a projective variety and L a line bundle which is globally generated but not ample. Then L defines a morphism $f_L: X \rightarrow Y \subseteq \mathbb P^n$ to some subvariety of projective space with $L = f_L^* H$, the pullback of the hyperplane class (which is very ample on Y). Now since L is not ample, the morphism $f_L$ must contract some curve on X: in other words, the Stein factorisation of $f_L$ consists of a nontrivial contraction followed by a finite morphism. (Here a contraction means a projective morphism with connected fibres.)

The moral is that for X any projective variety, globally generated but not ample line bundles on X are the "same thing" as nontrivial contraction morphisms $\pi: X \rightarrow Y$. (The correspondence can be made more precise by considering the face structure of the nef cone of X, but that's not really relevant to your question.)

Finally, a word on terminology: a line bundle L on X of which some power $L^k$ is globally generated is called (by some people at least) semi-ample. For some purposes this turns out to be a better notion than globally generated, just as ample is a better notion than very ample.

Not to appear ungrateful, but I suggest that you select another answer in place of mine, since mine addresses only a small part of one of your question. Those of e.g. VA and Andrea Ferretti are more complete.
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user5117May 24 '10 at 14:49

"select" -> "accept"
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user5117May 24 '10 at 14:49

Thanks. I had tried to accept multiple answers since I found them all helpful. I didn't realize that I had to select one 'accepted answer'.
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A. PascalMay 24 '10 at 19:19

I am having a senior moment on globally generated line bundles, so I'll leave this notion aside.

If $X$ is a non-hyperelliptic curve and $P,Q$ are distinct points on $X$, then ${\cal{O}}_X(2P-Q)$ is ample but has no global sections and is not very ample either. The canonical divisor on curves of genus at least two is very ample but has $H^1$.

Is a hard question in general, even the case of the canonical divisor (when it is ample) is difficult and important in dimension bigger than one. For a curve of genus $g$, the minimal $k$ depends on $L$ but if $k\deg L > g$, $L^k$ has sections and if $k\deg L > 2g$, $L^k$ is very ample. All this follows from Riemann-Roch.

I think the answer is no, but I can't think of a counterexample right now.

The answer to 3 is also yes (assuming X is projective): the morphism associated to a globally generated ample line bundle is necessarily finite, since it cannot contract any subvariety of positive dimension. Once you have a finite morphism to projective space, you can start projecting away from points as before.
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damianoMay 22 '10 at 22:28

1) Let $C$ be a curve (of genues at least 2) with a $2:1$ morhpism on $\mathbb{P}^1$. Then $K_C$ has positive degree, hence it is ample. But you can check by Riemann-Hurwitz that $K_C$ is just the pullback of some multiple (depending on the genus) of the hyperplane section on $\mathbb{P}^1$. Moreover by Riemann-Roch you see that all sections of $K_C$ are pulled back by $\mathbb{P}^1$. It follows easily that the map given by sections of $K_C$ is just the map to $\mathbb{P}^1$ followed by a Veronese embedding; in particular it is not injective. So $K_C$ is ample but not very ample. Moreover by Serre duality you have $h^1(K_C) = h^0(\mathcal{O}) = 1$.

For a very ample line bundle with cohomology EDIT: removed nonsense

2) There is a bound depending only on the dimension of $X$ and the coefficients of the Hilbert polynomial of $L$; this is known as Matsusaka big theorem.

3) In general no, since $L$ will not give you a finite map. But it may happen that $L$ gives a finite map which is not an isomorphism; in this case you can then project to get a finite map to a projective space.

For instance take the double covering of $\mathbb{P}^2$ ramified over a sexitc, say this is $\pi \colon X \to \mathbb{P}^2$; then the line bundle $\pi^{*}(\mathcal{O}(1))$ gives the map $\pi$ itself, which is finite. But is it true that $\pi^{*}(\mathcal{O}(1))$ is ample, for instance by Kleiman's criterion. More generally the pullback of an ample line bundle under a finite map is still ample.

ehehe, while we are at it, in 1) you should also assume that C has genus at least two.
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damianoMay 23 '10 at 0:48

You're right; I started with the phrase let C be a hyperelliptic curve, by which it is usually implicitly assumed that the genus is at least 2, and then modified the answer to write out what that means.
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Andrea FerrettiMay 23 '10 at 1:21